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I am trying to solve the following integral with trig substitutions. However, I get a different answer for two substitutions that should yield the same result. $$\int\frac{dx}{\sqrt{4-9x^2}}$$

  • For the first trig sub, I set $9x^2 = 4\cos^2\theta$. This simplifies to: $x = \frac{2}{3}\cos\theta$, and $dx = -\frac{2}{3}\sin\theta$. Substituting in, I get: $$\int\frac{-2\sin\theta}{6\sin\theta} = -\frac{\theta}{3} = -\frac{1}{3}\,\cos^{-1}\left(\frac{3x}{2}\right)+C \tag{1}$$

  • For the second trig sub, I set $9x^2 = 4\sin^2\theta$. This simplifies to: $x = \frac{2}{3}\sin\theta$, and $dx = \frac{2}{3}\cos\theta$. Substituting in, I get: $$\int\frac{2\cos\theta}{6\cos\theta} = \frac{\theta}{3} = \frac{1}{3}\,\sin^{-1}\left(\frac{3x}{2}\right)+C \tag{2}$$

My question is:

Why do these two trig substitutions yield different results graphically? Shouldn't they result in the same graph?

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1 Answer 1

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Since $(-\arcsin)'(x)=\arccos'(x)=-\dfrac1{\sqrt{1-x^2}}$, you got twice the same thing.

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  • $\begingroup$ However, when I graph these, they are not equal. Why is this? $\endgroup$
    – Jay
    Commented Mar 24, 2019 at 16:01
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    $\begingroup$ Because their difference is a constant:$$\arccos(x)+\arcsin(x)=\frac\pi2.$$So, your $C$'s are actually two distinct constants. $\endgroup$ Commented Mar 24, 2019 at 16:03
  • $\begingroup$ m.wolframalpha.com/input/?i=y%3D-%281%2F3%29+%5Carccos+%28x%29 .Perhaps helpful you have with (1/3)sin^{-1} (x) expression an additional constant -π/6 which you can absorb in C. $\endgroup$ Commented Mar 24, 2019 at 16:05

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